What is the vertex form of the equation of the parabola with a focus at (1,-9) and a directrix of #y=-1 #?

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Answer 1 · 2017-04-09T06:03:55

#y=-1/16(x-1)^2+5#

Parabola is the locus of a point which moves so that its distance from a point called focus and a line called directrix is always same.

Hence a point, say #(x,y)# on the desired parabola will be equidistant from focus #(1,-9)# and directrix #y=-1# or #y+1=0#.

As the distance from #(1,-9)# is #sqrt((x-1)^2+(y+9)^2)# and from #y+1# is #|y+1|#, we have

#(x-1)^2+(y+9)^2=(y+1)^2#

or #x^2-2x+1+y^2+18y+81=y^2+2y+1#

or #x^2-2x+16y+81=0#

or #16y=-1(x^2-2x+1-1)-81#

or #16y=--(x^2-2x+1)+1-81#

or #y=-1/16(x-1)^2+5#

Hence, vertex is #(1,-5)# and axis of symmetry is #x=1#

graph{(y+1/16(x-1)^2+5)(y+1)(x-1)((x-1)^2+(y+9)^2-0.04)=0 [-20.08, 19.92, -17.04, 2.96]}